zoukankan      html  css  js  c++  java
  • codeforces 915 C(思维+模拟)

    C. Permute Digits
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    You are given two positive integer numbers a and b. Permute (change order) of the digits of a to construct maximal number not exceeding b. No number in input and/or output can start with the digit 0.

    It is allowed to leave a as it is.

    Input

    The first line contains integer a (1 ≤ a ≤ 1018). The second line contains integer b (1 ≤ b ≤ 1018). Numbers don't have leading zeroes. It is guaranteed that answer exists.

    Output

    Print the maximum possible number that is a permutation of digits of a and is not greater than b. The answer can't have any leading zeroes. It is guaranteed that the answer exists.

    The number in the output should have exactly the same length as number a. It should be a permutation of digits of a.

    Examples
    input
    Copy
    123
    222
    output
    Copy
    213
    input
    Copy
    3921
    10000
    output
    Copy
    9321
    input
    Copy
    4940
    5000
    output
    Copy
    4940

     题意:给你两个整数a,b。让你把组成a的数字排列后成比b小的最大的数字。

    思路:最开始想,从b的最高位(最左面的数)遍历,从a中找相等的数,当找到的没有相等,而是小于,则把a中没有输出的数从大到小输出(反解:98 ,91)

        后来想dfs跑,但是会超时(例如200000001,200000000))

    正解:把a,b拆出来到数组中,把a从大到小排序。从b的最高位(最左面的数)遍历,从a中找相等的数,当找到相等则取,如果没有相等的情况还存在两种情况,只存在大于的b[i]的数,和只存在小于b[i]的数

        如果后者,则把a中没有输出的数从大到小输出。如果存在前者,则回溯(i -=  2),直到找到一个可以找到有比b[i](这个i和前面的i不同),然后把a中没有输出的数从大到小输出即可。

    #include <iostream>
    #include <cmath>
    #include <cstdio>
    #include <cstring>
    #include <string>
    #include <map>
    #include <iomanip>
    #include <algorithm>
    #include <queue>
    #include <stack>
    #include <set>
    #include <vector>
    //const int maxn = 1e5+5;
    #define ll long long
    #define inf 0x3f
    #define FOR(i,a,b) for( int i = a;i <= b;++i)
    #define bug cout<<"--------------"<<endl
    #define bugs cout<<"SSSSSSSSSSSSSSS"<<endl
    
    ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
    ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
    const int maxn = 5100;
    using namespace std;
    int a[50],b[50],vis[50],cun[50];
    int t[50];
    int tail1,tail2;
    
    int n;
    int findsame(int k)
    {
        for(int i = 1;i <= n; ++i)
        {
            if(a[i] == b[k] && vis[i] == 0) return i;
        }
        return 0;
    }
    int findsmall(int k)
    {
        for(int i = 1;i <= n; ++i)
        {
            if(a[i] < b[k] && vis[i] == 0) return i;
        }
        return 0;
    }
    int main()
    {
        
        //freopen("C:\ACM\input.txt","r",stdin);
        ll x,y;
        cin>>x>>y;
         tail1 = 0;
         tail2 = 0;
        while(x)
        {
            a[++tail1] = x % 10;
            x /= 10;
        }
        while(y)
        {
            b[++tail2] = y % 10;
            y /= 10;
        }    
        sort(a+1,a+tail1+1);
        reverse(a+1,a+1+tail1);
        reverse(b+1,b+1+tail2);
    
        if(tail1 < tail2 )
        {
            for(int i = 1;i <= tail1; ++i)
                cout<<a[i];
        }
        else 
        {
            n = tail1;
            int flag = 0;
            int Head = 0;
            stack<int>sta;
            for(int i = 1;i <= n; ++i)
            {
    
                if(flag == 1)
                {
                    int temp = sta.top();sta.pop();
                    vis[temp] = 0;
                    if(findsmall(i) != 0)
                    {
                        
                        cun[i] = a[findsmall(i)];
                        Head = i;
                        vis[findsmall(i)] = 1;
                        break;
                    } 
                    else i-=2;
                    continue;
                }
                if(findsame(i) != 0)
                {
                    cun[i] = a[findsame(i)];
                    sta.push(findsame(i));
                    //cout<<findsame(i)<<endl;
                    
                    vis[findsame(i)] = 1; 
                    
                    
                }
                else if(findsmall(i) != 0)
                {
                    cun[i] = a[findsmall(i)];
                    sta.push(findsmall(i));
    
    
                    Head = i ;
                    vis[findsmall(i)] = 1;
                    
                    break;
                }
                else 
                {
                    i -= 2;
                    flag = 1;
                }
                if(i == n) Head = n;
            }
    
    
    /*        for(int i = 1;i <= n; ++i) cout<<vis[i]<<" ";
            cout<<endl;
            bug;*/
            for(int i = 1;i <= Head; ++i) cout<<cun[i]; 
            for(int i = 1;i <= n; ++i) if(vis[i] ==0) cout<<a[i];
    
        }
    }

      

  • 相关阅读:
    windows 创建python独立开发环境
    sql多列排序
    mysql 导入sql脚本中文乱码问题
    廖雪峰Python教学课后作业---datetime
    poj 1004:Financial Management(水题,求平均数)
    【POJ水题完成表】
    poj 1003:Hangover(水题,数学模拟)
    ytu 2558: 游起来吧!超妹!(水题,趣味数学题)
    poj 1005:I Think I Need a Houseboat(水题,模拟)
    hdu 2393:Higher Math(计算几何,水题)
  • 原文地址:https://www.cnblogs.com/jrfr/p/11525476.html
Copyright © 2011-2022 走看看